Derivation of voltage for vacuum diode space charge region

Nuahaun
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I have found the 1D differential equation relating voltage and position for a vacuum diode in the space charge region, which is

\frac{d^2V}{dx^2} = constant * V^{-1/2}

and I know the solution to be

V(x) = V_0 \left(\frac{x}{d}\right)^{4/3}

which is found by multiplying both sides by V' = \frac{dV}{dx} and then integrating the following expression with homogeneous boundary conditions:

\int V' dV' = constant*\int V^{-1/2} dV

What I don't understand is why this trick is even necessary. As far as I can tell the differential equation can be solved by a simple separation of variables, which gives an answer of

V(x) = V_0 \left(\frac{x}{d}\right)^{4/5}

The two answers are different so obviously it's a mistake to separate variables, but for the life of me I can't tell where it is. Could anyone enlighten me? Thanks.
 
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Nuahaun said:
What I don't understand is why this trick is even necessary. As far as I can tell the differential equation can be solved by a simple separation of variables, which gives an answer of

I'm not exactly sure what you mean by "separation of variables" in this case. To apply separation of variables, V would have to be a function of more than one variable; but in this case, it is a function of only x. SOV is typically used to turn PDEs into ODEs, not to solve an ODE.

I can only guess that what you meant to say was that you tried treating it as a "separable differential equation"; but you have a second order DE...how exactly does one integrate V^{1/2} d^2 V and dx^2?
 
yeah wasn't totally sure what was meant, however a trick to reduce it to 2 first order DEs as follows:

First write
P = V' = \frac{dV}{dx}
then
P' = V" = \frac{dp}{dx} = \frac{dp}{dV}.\frac{dV}{dx} = p.\frac{dp}{dV}

otherwise just assume solutions V~xy diff & equate coeffs
 
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Thanks for your replies. My mistake was that I though I could solve V^{1/2} d^2 V = dx^2, which stemmed from an incomplete understanding of the method of separation of variables (I took the fact that you could solve something like d^2x/dt^2= F/m using double integration, and incorrectly applied that to the differential equation above). I reviewed my notes, and now see why that isn't possible.
 
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